Prove-by-mathematical-induction-r-1-n-1-r-r-1-n-n-1-

Question Number 157351 by physicstutes last updated on 22/Oct/21

Prove by mathematical induction

\underset{r = 1}{\sum^{n}} \frac{1}{r (r + 1)} = \frac{n}{n + 1}

Commented by hknkrc46 last updated on 22/Oct/21

★ \frac{1}{r (r + 1)} = \frac{(r + 1) - r}{r (r + 1)}

= \frac{r + 1}{r (r + 1)} - \frac{r}{r (r + 1)}

= \frac{1}{r} - \frac{1}{r + 1}

★ \underset{r = 1}{\sum^{n}} \frac{1}{r (r + 1)} = \underset{r = 1}{\sum^{n}} (\frac{1}{r} - \frac{1}{r + 1})

= (\frac{1}{1} - \frac{1}{2}) + (\frac{1}{2} - \frac{1}{3}) + \cdot \cdot \cdot + (\frac{1}{n} - \frac{1}{n + 1})

= \frac{1}{1} - \frac{1}{n + 1} = \frac{n}{n + 1}

Answered by ajfour last updated on 22/Oct/21

S = \underset{r = 1}{\sum^{n}} \frac{1}{T_{r}} = \underset{r = 1}{\sum^{n}} (\frac{1}{r} - \frac{1}{r + 1})

= \underset{r = 1}{\sum^{n}} (t_{r} - t_{r + 1}) = 1 - \frac{1}{n + 1}

\Rightarrow S = \frac{n}{n + 1}

Answered by physicstutes last updated on 22/Oct/21

∙ prove for n = 1 .

LHS = \underset{r = 1}{\sum^{1}} \frac{1}{r (r + 1)} = \frac{1}{1 (1 + 1)} = \frac{1}{2}

RHS = \frac{1}{1 + 1} = \frac{1}{2}

\Rightarrow true for n = 1 .

∙ Assume it is true for n = k

\Rightarrow \underset{r = 1}{\sum^{k}} \frac{1}{r (r + 1)} = \frac{k}{k + 1}

∙ Prove for n = k + 1 .

\underset{r = 1}{\sum^{k + 1}} \frac{1}{r (r + 1)} = \underset{r = 1}{\sum^{k}} \frac{1}{r (r + 1)} + \frac{1}{(k + 1) (k + 2)}

= \frac{k}{k + 1} + \frac{1}{(k + 1) (k + 2)}

= \frac{1}{k + 1} (k + \frac{1}{k + 2})

= \frac{1}{k + 1} (\frac{k^{2} + 2 k + 1}{k + 2})

= \frac{1}{k + 1} (\frac{{(k + 1)}^{2}}{k + 2})

= \frac{k + 1}{k + 2}

\Rightarrow true for n = k + 1

hence true \forall n \in Z

Answered by som(math1967) last updated on 22/Oct/21

T o p r o v e

\frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \frac{1}{3 \times 4} + \dots + \frac{1}{n (n + 1)} = \frac{n}{n + 1}

p (1) L . H . S = \frac{1}{1 \times 2} = \frac{1}{2}

R . H . S = \frac{1}{1 + 1} = \frac{1}{2}

∴ t r u e f o r p (1)

l e t t r u e f o r p (m)

∴ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{m (m + 1)} = \frac{m}{m + 1}

n o w p (m + 1)

= \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{m (m + 1)} + \frac{1}{(m + 1) (m + 2)}

= \frac{m}{m + 1} + \frac{1}{(m + 1) (m + 2)} ★

= \frac{m (m + 2) + 1}{(m + 1) (m + 2)}

= \frac{{(m + 1)}^{2}}{(m + 1) (m + 2)} = \frac{m + 1}{m + 2}

∴ t r u e f o r p (m + 1)

∴ \underset{r = 1}{\sum^{n}} \frac{1}{r (r + 1)} = \frac{n}{n + 1}

★ \frac{1}{1 \times 2} + \frac{1}{2 \times 3} + \dots + \frac{1}{m (m + 1)} = \frac{m}{m + 1}

Commented by peter frank last updated on 22/Oct/21

great